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I tried to find an expression that fits the power following serie but didn't succeed, it looks almost like the derivative of arcsin, but ...no, it's not:

$$f(x)=\sum_{n=0}^\infty \frac{x^n(2n)!}{n!}$$

Of course $-1<x<1$ Maybe it's well known, but I didn't find it, neither by integration nor derivation, nor simply searching. I guess it doesn't exist but I'd like a confirmation.

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    $\begingroup$ This sum diverges for all $x\ne 0$. $\endgroup$ – Frpzzd Aug 10 '17 at 23:56
  • $\begingroup$ yep, I asked too fast, that's pretty obvious now you tell me. $\endgroup$ – kalish Oct 5 '17 at 21:05
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This is bad.

Let $$a_n=\frac{x^n(2n)!}{n!}$$ Then $$\frac{a_{n+1}}{a_n}=\frac{\frac{x^{n+1}(2n+2)!}{(n+1)!}}{\frac{x^n(2n)!}{n!}}$$ $$\frac{a_{n+1}}{a_n}=\frac{x(2n+1)(2n+2)}{n+1}$$ and so, no matter what $x$ is (except $x=0$), $$\lim_{n\to\infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|=\lim_{n\to\infty} \bigg|\frac{x(2n+1)(2n+2)}{n+1}\bigg |=\color{red}{\infty}$$ And so the sum diverges for all nonzero values of $x$.

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    $\begingroup$ Bleh, who needs convergence? $\endgroup$ – Simply Beautiful Art Aug 11 '17 at 0:19
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    $\begingroup$ yeah thanks, I might have been drunk that day. I don't remember why I needed it. $\endgroup$ – kalish Oct 5 '17 at 21:06
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    $\begingroup$ @kalish If you think this or another answer has answered your question, please click the green $\color{green}\checkmark$ to the side to accept the answer. $\endgroup$ – Simply Beautiful Art Oct 5 '17 at 22:55
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As Nilknarf shows, the sum diverges, however, it is possible to obtain a Mittag-Leffler sum.

We will need to require $\Re(x)\le0$ and we get

\begin{align}\mathcal M_2\sum_{n=0}^\infty\frac{(2n)!}{n!}x^n&=\int_0^\infty e^{-t}\left(\sum_{n=0}^\infty\frac1{n!}x^nt^{2n}\right)~\mathrm dt\\&=\int_0^\infty\exp(xt^2-t)~\mathrm dt\\&=\frac{\pi^{1/2}}{2e^{1/4x}\sqrt{-x}}\left(1-\operatorname{erf}\left(\frac1{2\sqrt{-x}}\right)\right)\end{align}

So I suppose, if it were anything, it'd be that, where $\operatorname{erf}(z)$ is the error function.

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  • $\begingroup$ Bugger me I didn't even check the radius and don't remember why I needed that expression, but it is sure it appeared in a physical situation. Thanks all of you for your kind answers. Edit: now I remember, and it was about integrating a gaussian or something related to a gaussian, so thanks. $\endgroup$ – kalish Oct 5 '17 at 21:04

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